<span>mainly the interaction of analyse and the base comes in mind and that way how solvable it is. if you but ethanol in water there is a change that it's blocking the space between the base and water and so on also let's other substances be carried away. so basically it would give lower distance for more polar and vice versa. also what kind of base are you using if it's selective for some substances </span>
Answer:
Explanation:
H3PO4(aq) + 3NaOH(aq) → Na3PO4(aq) + 3H2O(l)
mole of NaOH = 23.6 * 10 ⁻³L * 0.2M
= 0.00472mole
let x be the no of mole of H3PO4 required of 0.00472mole of NaOH
3 mole of NaOH required ------- 1 mole of H3PO4
0.00472mole of NaOH ----------x
cross multiply
3x = 0.0472
x = 0.00157mole
[H3PO4] = mole of H3PO4 / Vol. of H3PO4
= 0.00157mole / (10*10⁻³l)
= 0.157M
<h3>The concentration of unknown phosphoric acid is 0.157M</h3>
Answer:
ΔHr = -275 kj
Explanation:
It is possible to obtain the net change in enthalpy for the formation of one mole of lead(II) sulfate from lead, lead(IV) oxide, and sulfuric acid using the reactions:
(1) H₂SO₄(l) → SO₃(g) + H₂O (l) ΔH=+113kJ
(2) Pb(s) + PbO₂(s) + 2SO₃(g) → 2PbSO₄(s) ΔH=−775kJ
If you sum (1) + ¹/₂(2) you will obtain:
H₂SO₄(l) + ¹/₂Pb(s) + ¹/₂PbO₂(s) → PbSO₄(s) + H₂O(l)
Using Hess's law, the net change in enthalpy for this reaction could be obtained as:
ΔHr = ΔH(1) + ¹/₂ΔH(2)
ΔHr = +113kJ + ¹/₂ -775kJ
ΔHr = -275 kJ
Answer:
Explanation:
6.90 mol x grams
2As + 6NaOH = 2Na3AsO3 + 3H2
6 mol 2 mol
192 g/mol
6.90 mol NaOH x 2 mol Na3AsO3/6 mol NaOH== 2.3 mol Na3AsO3
2.3 mol Na3AsO3 x 192 g/mol = 442 g Na3AsO3